Closed form for a conditional trigonometric series

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Do you know an easy way to prove the following $forall~L$?
$sumlimits_scriptstyle k = 1,~k ne Latop
scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
l) right) - 2sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
- L) right) + 1 = left{ beginarrayl
(N - 2)^2rm~~if~n = 0\
4rm~~~if~n ne 0
endarray right.$



All parameters are integer numbers.







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  • Incorporate back the terms associated to $k=L$ and $l=L$, then exploit $costheta=textRe(e^itheta)$ and the formula for the sum of geometric progressions.
    – Jack D'Aurizio♦
    Aug 10 at 20:52










  • Thank you for your comments. I like Jack D'Aurizio 's approach. This question comes from here: https://math.stackexchange.com/questions/2878067/proof-that-certain-sums-of-complex-numbers-in-a-unitary-matrix-produce-the-sam
    – HS TQ
    Aug 10 at 21:02















up vote
0
down vote

favorite












Do you know an easy way to prove the following $forall~L$?
$sumlimits_scriptstyle k = 1,~k ne Latop
scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
l) right) - 2sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
- L) right) + 1 = left{ beginarrayl
(N - 2)^2rm~~if~n = 0\
4rm~~~if~n ne 0
endarray right.$



All parameters are integer numbers.







share|cite|improve this question




















  • Incorporate back the terms associated to $k=L$ and $l=L$, then exploit $costheta=textRe(e^itheta)$ and the formula for the sum of geometric progressions.
    – Jack D'Aurizio♦
    Aug 10 at 20:52










  • Thank you for your comments. I like Jack D'Aurizio 's approach. This question comes from here: https://math.stackexchange.com/questions/2878067/proof-that-certain-sums-of-complex-numbers-in-a-unitary-matrix-produce-the-sam
    – HS TQ
    Aug 10 at 21:02













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Do you know an easy way to prove the following $forall~L$?
$sumlimits_scriptstyle k = 1,~k ne Latop
scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
l) right) - 2sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
- L) right) + 1 = left{ beginarrayl
(N - 2)^2rm~~if~n = 0\
4rm~~~if~n ne 0
endarray right.$



All parameters are integer numbers.







share|cite|improve this question












Do you know an easy way to prove the following $forall~L$?
$sumlimits_scriptstyle k = 1,~k ne Latop
scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
l) right) - 2sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
- L) right) + 1 = left{ beginarrayl
(N - 2)^2rm~~if~n = 0\
4rm~~~if~n ne 0
endarray right.$



All parameters are integer numbers.









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 10 at 20:40









HS TQ

375




375











  • Incorporate back the terms associated to $k=L$ and $l=L$, then exploit $costheta=textRe(e^itheta)$ and the formula for the sum of geometric progressions.
    – Jack D'Aurizio♦
    Aug 10 at 20:52










  • Thank you for your comments. I like Jack D'Aurizio 's approach. This question comes from here: https://math.stackexchange.com/questions/2878067/proof-that-certain-sums-of-complex-numbers-in-a-unitary-matrix-produce-the-sam
    – HS TQ
    Aug 10 at 21:02

















  • Incorporate back the terms associated to $k=L$ and $l=L$, then exploit $costheta=textRe(e^itheta)$ and the formula for the sum of geometric progressions.
    – Jack D'Aurizio♦
    Aug 10 at 20:52










  • Thank you for your comments. I like Jack D'Aurizio 's approach. This question comes from here: https://math.stackexchange.com/questions/2878067/proof-that-certain-sums-of-complex-numbers-in-a-unitary-matrix-produce-the-sam
    – HS TQ
    Aug 10 at 21:02
















Incorporate back the terms associated to $k=L$ and $l=L$, then exploit $costheta=textRe(e^itheta)$ and the formula for the sum of geometric progressions.
– Jack D'Aurizio♦
Aug 10 at 20:52




Incorporate back the terms associated to $k=L$ and $l=L$, then exploit $costheta=textRe(e^itheta)$ and the formula for the sum of geometric progressions.
– Jack D'Aurizio♦
Aug 10 at 20:52












Thank you for your comments. I like Jack D'Aurizio 's approach. This question comes from here: https://math.stackexchange.com/questions/2878067/proof-that-certain-sums-of-complex-numbers-in-a-unitary-matrix-produce-the-sam
– HS TQ
Aug 10 at 21:02





Thank you for your comments. I like Jack D'Aurizio 's approach. This question comes from here: https://math.stackexchange.com/questions/2878067/proof-that-certain-sums-of-complex-numbers-in-a-unitary-matrix-produce-the-sam
– HS TQ
Aug 10 at 21:02











1 Answer
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for $n=0$ we have
$$ (N-1)^2-2(N-1)+1 =(N-2)^2 $$



If $nne0$ , using the symmetry $ cos(pi - x) = -cos(x)$ (or any other method) you should be able to convince yourself that
$$ sumlimits_k = 1^N cos left( frac2pi Nn(k
- L) right) =0
$$
For all integers $N, L, n ne 0$.



So
$$sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
- L) right) = 0 - cos left( frac2pi Nn(L
- L) right) =-1
$$
And
$$sumlimits_scriptstyle k = 1,~k ne Latop
scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
l) right)
\ = sumlimits_scriptstyle k = 1,~k ne L^N - cos left( frac2pi Nn(k -
L) right)
\ = cos left( frac2pi Nn(L
- L) right)
=1$$






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    for $n=0$ we have
    $$ (N-1)^2-2(N-1)+1 =(N-2)^2 $$



    If $nne0$ , using the symmetry $ cos(pi - x) = -cos(x)$ (or any other method) you should be able to convince yourself that
    $$ sumlimits_k = 1^N cos left( frac2pi Nn(k
    - L) right) =0
    $$
    For all integers $N, L, n ne 0$.



    So
    $$sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
    - L) right) = 0 - cos left( frac2pi Nn(L
    - L) right) =-1
    $$
    And
    $$sumlimits_scriptstyle k = 1,~k ne Latop
    scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
    l) right)
    \ = sumlimits_scriptstyle k = 1,~k ne L^N - cos left( frac2pi Nn(k -
    L) right)
    \ = cos left( frac2pi Nn(L
    - L) right)
    =1$$






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      for $n=0$ we have
      $$ (N-1)^2-2(N-1)+1 =(N-2)^2 $$



      If $nne0$ , using the symmetry $ cos(pi - x) = -cos(x)$ (or any other method) you should be able to convince yourself that
      $$ sumlimits_k = 1^N cos left( frac2pi Nn(k
      - L) right) =0
      $$
      For all integers $N, L, n ne 0$.



      So
      $$sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
      - L) right) = 0 - cos left( frac2pi Nn(L
      - L) right) =-1
      $$
      And
      $$sumlimits_scriptstyle k = 1,~k ne Latop
      scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
      l) right)
      \ = sumlimits_scriptstyle k = 1,~k ne L^N - cos left( frac2pi Nn(k -
      L) right)
      \ = cos left( frac2pi Nn(L
      - L) right)
      =1$$






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        for $n=0$ we have
        $$ (N-1)^2-2(N-1)+1 =(N-2)^2 $$



        If $nne0$ , using the symmetry $ cos(pi - x) = -cos(x)$ (or any other method) you should be able to convince yourself that
        $$ sumlimits_k = 1^N cos left( frac2pi Nn(k
        - L) right) =0
        $$
        For all integers $N, L, n ne 0$.



        So
        $$sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
        - L) right) = 0 - cos left( frac2pi Nn(L
        - L) right) =-1
        $$
        And
        $$sumlimits_scriptstyle k = 1,~k ne Latop
        scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
        l) right)
        \ = sumlimits_scriptstyle k = 1,~k ne L^N - cos left( frac2pi Nn(k -
        L) right)
        \ = cos left( frac2pi Nn(L
        - L) right)
        =1$$






        share|cite|improve this answer












        for $n=0$ we have
        $$ (N-1)^2-2(N-1)+1 =(N-2)^2 $$



        If $nne0$ , using the symmetry $ cos(pi - x) = -cos(x)$ (or any other method) you should be able to convince yourself that
        $$ sumlimits_k = 1^N cos left( frac2pi Nn(k
        - L) right) =0
        $$
        For all integers $N, L, n ne 0$.



        So
        $$sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
        - L) right) = 0 - cos left( frac2pi Nn(L
        - L) right) =-1
        $$
        And
        $$sumlimits_scriptstyle k = 1,~k ne Latop
        scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
        l) right)
        \ = sumlimits_scriptstyle k = 1,~k ne L^N - cos left( frac2pi Nn(k -
        L) right)
        \ = cos left( frac2pi Nn(L
        - L) right)
        =1$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 10 at 21:30









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